Work and Energy

CIE A-Level Further Mathematics· 30 min read

1. Work Done by Constant and Variable Forces★★☆☆☆⏱ 15 min

📘 Definition

Work Done

W

The energy transferred when a force acts on an object moving through a displacement. For a force $F$ over displacement element $d r$ , work done is given by the line integral: $W = \int F \cdot d r$

Example:

A 10 N constant force moving an object 5 m in the direction of the force does $10 \times 5 = 50$ J of work.

For constant forces, the line integral simplifies to $W = F d cos θ$ , where $θ$ is the angle between the force vector and displacement. If the force is perpendicular to displacement, $cos θ = 0$ so work done equals zero. For variable forces acting along a straight line, this reduces to $W = \int_{x_{1}}^{x_{2}} F (x) d x$ .

📐 Worked Example

A variable force $F (x) = 3 x^{2} + 2 x$ (in newtons) acts on a particle moving along the x-axis from $x = 0$ to $x = 3$ m. Calculate the total work done by the force.

1
Use the formula for work done by a 1D variable force:
2
$W = \int_0^3 (3x^2 + 2x) \mathrm{d}x$
3
Integrate term-by-term:
4
$\left[ x^3 + x^2 \right]_0^3$
5
Substitute the upper and lower bounds:
6
$(3^3 + 3^2) - (0 + 0) = 27 + 9 = 36$
7
Total work done is 36 J.

Exam tip:

Always check the direction of the force relative to displacement — negative work indicates energy is removed from the system (e.g. work done by friction).

2. Work-Energy Principle★★★☆☆⏱ 20 min

📘 Definition

Work-Energy Principle

The net work done on a particle by all external forces equals the change in the kinetic energy of the particle: $W_{net} = Δ E_{k} = E_{k final} - E_{k initial}$

Example:

If a particle loses 20 J of kinetic energy, the net work done on it by all forces is -20 J.

This principle holds for both constant and variable forces, and can be extended to systems of particles. When only conservative forces (gravity, elastic forces) do work, total mechanical energy is conserved: $E_{k} + E_{p} + E_{e} = constant$ . If non-conservative forces like friction act, we account for work done against these forces in our energy balance.

📐 Worked Example

A 2 kg particle is projected up a rough inclined plane with initial speed 10 m/s. The plane is inclined at 30° to the horizontal, and the coefficient of friction is 0.2. Use the work-energy principle to find the maximum distance travelled up the plane. Take $g = 9.8$ m/s².

1
Let maximum distance be $s$ . At maximum distance, final speed is 0, so change in kinetic energy is:
2
$\Delta E_k = 0 - \frac{1}{2}mv^2 = -\frac{1}{2}(2)(10)^2 = -100 \text{ J}$
3
Calculate work done by each force: Normal reaction is perpendicular to displacement so work done is 0. Work done by gravity and friction are both negative:
4
$W_g = -mg s \sin 30^\circ = -9.8s, \quad W_f = -\mu mg \cos 30^\circ s = -3.39s$
5
Apply the work-energy principle: total work done equals change in kinetic energy:
6
$-9.8s - 3.39s = -100 \implies 13.19s = 100$
7
Solve for $s$ :
8
$s \approx 7.6 \text{ m}$

3. Elastic Potential Energy★★★☆☆⏱ 20 min

When an elastic string or spring is stretched or compressed from its natural length, it stores elastic potential energy equal to the work done to deform it from its natural length. This is a core concept for problems involving elastic systems in CIE exams.

📘 Definition

Elastic Potential Energy

E_e

Energy stored in a deformed elastic material, given by $E_{e} = \frac{1}{2} k x^{2} = \frac{λ x ^{2}}{2 l _{0}}$ , where $k$ is the spring constant, $x$ is extension/compression, $λ$ is modulus of elasticity and $l_{0}$ is natural length.

Example:

A spring with $k = 10$ N/m stretched by 2 m stores $\frac{1}{2} (10) (2)^{2} = 20$ J of elastic energy.

📐 Worked Example

A light elastic string has natural length 2 m and modulus of elasticity 40 N. One end is fixed to a point on a rough horizontal table, the other end attached to a 3 kg particle. The particle is pulled 4 m from the fixed end and released from rest. Find its speed when it returns to the natural length position, given $μ = 0.2$ and $g = 9.8$ m/s².

1
Calculate spring constant $k = λ / l_{0} = 40/2 = 20$ N/m. Initial extension is $x = 4 - 2 = 2$ m.
2
Work done against friction over 2 m is:
3
$W_{\text{friction}} = \mu mg \times 2 = 0.2 \times 3 \times 9.8 \times 2 = 11.76 \text{ J}$
4
Apply energy conservation: initial elastic potential energy equals final kinetic energy plus work done against friction:
5
$\frac{1}{2} k x^2 = \frac{1}{2} m v^2 + 11.76$
6
Substitute values:
7
$40 = 1.5 v^2 + 11.76 \implies 1.5 v^2 = 28.24$
8
Solve for $v$ :
9
$v \approx 4.34 \text{ m/s}$

Exam tip:

Always calculate the change in elastic potential energy, not just the final value, when the spring starts from a non-zero extension.

4. Power and Efficiency★★★☆☆⏱ 15 min

Power is the rate of doing work, and is commonly tested in problems involving vehicles moving with constant engine power, which cannot be solved directly with constant acceleration kinematics.

📘 Definition

Power & Efficiency

Instantaneous power is $P = F v$ for a force $F$ acting on an object moving at instantaneous speed $v$ in the direction of the force. Average power is $P = W / t$ . Efficiency $η = \frac{Useful output power}{Input power} \times 100%$ .

📐 Worked Example

A 1200 kg car moves up a hill inclined at $arcsin (1/10)$ . The engine produces constant power of 30 kW, and resistance to motion is 200 N. Find the acceleration when speed is 10 m/s. Take $g = 9.8$ m/s².

1
Find driving force from power: $F = P / v$
2
$F = 30000 / 10 = 3000 \text{ N}$
3
Component of weight down the hill:
4
$mg \sin\theta = 1200 \times 9.8 \times 1/10 = 1176 \text{ N}$
5
Apply Newton's second law up the hill:
6
$F - \text{resistance} - mg \sin\theta = ma$
7
$3000 - 200 - 1176 = 1200a \implies 1624 = 1200a$
8
Solve for acceleration:
9
$a \approx 1.35 \text{ m/s}^2$

5. Common Pitfalls

Wrong move:

Forgetting the angle between force and displacement, using $W = F d$ for any force.

Why:

Work only depends on the component of force in the direction of movement, not the total magnitude of the force.

Correct move:

Always use $W = F d cos θ$ to account for the angle between force and displacement.

Wrong move:

Using modulus of elasticity directly in the elastic potential energy formula as $E_{e} = \frac{1}{2} λ x^{2}$ .

Why:

CIE questions usually give modulus of elasticity $λ$ , not spring constant $k$ . The formula depends on natural length.

Correct move:

Use $E_{e} = \frac{λ x ^{2}}{2 l _{0}}$ where $l_{0}$ is the natural length of the string or spring.

Wrong move:

Assuming mechanical energy is always conserved.

Why:

Energy conservation only holds when no non-conservative forces (like friction) do work.

Correct move:

Always add work done against non-conservative forces to the energy balance equation when friction is present.

Wrong move:

Using $P = F v$ to calculate average power over a journey.

Why:

$P = F v$ gives instantaneous power for the instantaneous speed $v$ , not average power over time.

Correct move:

Use $P = \frac{Total work done}{Total time}$ for average power calculations.

Wrong move:

Forgetting that internal forces can do work in systems of connected particles.

Why:

The work-energy principle for systems accounts for work done by both external and internal forces.

Correct move:

Always check for tension in inextensible strings, which can do work on individual connected particles.

6. Quick Reference Cheatsheet

Quantity	Formula	Notes
Work (constant force)	$W = F d cos θ$	$θ$ = angle between $F$ and displacement
Work (1D variable force)	$W = \int_{x_{1}}^{x_{2}} F (x) d x$
Work (2D variable force)	$W = \int F \cdot d r$
Work-Energy Principle	$W_{net} = Δ E_{k}$	Net work = change in kinetic energy
Elastic Potential Energy	$E_{e} = \frac{λ x ^{2}}{2 l _{0}} = \frac{1}{2} k x^{2}$	$x$ = extension from natural length
Energy Conservation	$E_{1} = E_{2} + W_{lost}$	$W_{lost}$ = work against non-conservative forces
Instantaneous Power	$P = F v$	$F$ in direction of instantaneous velocity $v$
Average Power	$P = \frac{W}{t}$	$W$ = total work done over time $t$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Variable force work done problem
2023 · 2
Elastic energy and work against friction
2024 · 1
Power on an inclined plane

Going deeper

syllabusCIE 9231 Further Mechanics Syllabus

What's Next

Work and energy is a foundational concept for all Further Mechanics topics, and regularly appears in combination with collisions, circular motion and rigid body dynamics in CIE 9231 exams. Mastering energy methods allows you to solve complex problems much faster than using Newton's second law directly, especially for variable force problems that are tricky with other approaches. These concepts will directly underpin your understanding of momentum and collisions, as well as gravitational potential energy in orbital mechanics problems. Use the links below to continue building your knowledge of Further Mechanics for the exam.